package leetcode101.data_structure;

/**
 * @author Synhard
 * @version 1.0
 * @class Code9
 * @description 23. 合并K个升序链表
 * 给你一个链表数组，每个链表都已经按升序排列。
 * <p>
 * 请你将所有链表合并到一个升序链表中，返回合并后的链表。
 * 示例 1：
 * <p>
 * 输入：lists = [[1,4,5],[1,3,4],[2,6]]
 * 输出：[1,1,2,3,4,4,5,6]
 * 解释：链表数组如下：
 * [
 * 1->4->5,
 * 1->3->4,
 * 2->6
 * ]
 * 将它们合并到一个有序链表中得到。
 * 1->1->2->3->4->4->5->6
 * 示例 2：
 * <p>
 * 输入：lists = []
 * 输出：[]
 * 示例 3：
 * <p>
 * 输入：lists = [[]]
 * 输出：[]
 * <p>
 * <p>
 * 提示：
 * <p>
 * k == lists.length
 * 0 <= k <= 10^4
 * 0 <= lists[i].length <= 500
 * -10^4 <= lists[i][j] <= 10^4
 * lists[i] 按 升序 排列
 * lists[i].length 的总和不超过 10^4
 * @tel 13001321080
 * @email 823436512@qq.com
 * @date 2021-05-10 9:00
 */
public class Code9 {

    static class ListNode {
        int val;
        ListNode next;

        ListNode() {
        }

        ListNode(int val) {
            this.val = val;
        }

        ListNode(int val, ListNode next) {
            this.val = val;
            this.next = next;
        }
    }

    public static void main(String[] args) {

    }

    public ListNode mergeKLists(ListNode[] lists) {
        if (lists.length == 0) {
            return null;
        }
        return merge(lists, 0, lists.length - 1);
    }

    private static ListNode merge(ListNode[] lists, int low, int high) {
        if (low == high) {
            return lists[low];
        }
        int mid = (low + high) >> 1;
        ListNode l1 = merge(lists, low, mid);
        ListNode l2 = merge(lists, mid + 1, high);
        return mergeTwoLists(l1, l2);
    }

    private static ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null || l2 == null) {
            return l1 == null ? l2 : l1;
        }
        if (l1.val < l2.val) {
            l1.next = mergeTwoLists(l1.next, l2);
            return l1;
        }
        if (l1.val > l2.val) {
            ListNode head = mergeTwoLists(l2.next, l1);
            l2.next = head;
            return l2;
        }
        ListNode head = mergeTwoLists(l1.next, l2.next);
        l1.next = l2;
        l2.next = head;
        return l1;
    }
}
